If A = begin{bmatrix} 0 & 1 & 2 1 & 2 & 3 3 | vedclass

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(A) We know that $A A^{-1} = I$,where $I$ is the identity matrix.Calculating the determinant $|A|$:$|A| = 0(2 - 3a) - 1(1 - 9) + 2(a - 6) = 0 + 8 + 2a

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$a = 1, c = -1$

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(A) We know that $A A^{-1} = I$,where $I$ is the identity matrix.Calculating the determinant $|A|$:$|A| = 0(2 - 3a) - 1(1 - 9) + 2(a - 6) = 0 + 8 + 2a - 12 = 2a - 4 = 2(a - 2)$.Since $A^{-1}$ exists,$|A| 
eq 0$,so $a 
eq 2$.The element at position $(1, 1)$ of $A^{-1}$ is $\frac{C_{11}}{|A|}$,where $C_{11}$ is the cofactor of $A_{11}$.$C_{11} = (2 	imes 1 - 3 	imes a) = 2 - 3a$.Given $A^{-1}_{11} = 1/2$,we have $\frac{2 - 3a}{2(a - 2)} = 1/2$.$2 - 3a = a - 2 \implies 4a = 4 \implies a = 1$.Now,to find $c$,we look at $A^{-1}_{23}$. This is $\frac{C_{32}}{|A|}$.$C_{32} = - \begin{vmatrix} 0 & 2 \ 1 & 3 \end{vmatrix} = -(0 - 2) = 2$.So,$c = \frac{2}{2(a - 2)} = \frac{1}{a - 2}$.Substituting $a = 1$,$c = \frac{1}{1 - 2} = -1$.Thus,$a = 1$ and $c = -1$.

If $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & c \\ 5/2 & -3/2 & 1/2 \end{bmatrix}$,then:

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